Below is the syntax highlighted version of Permutations.java
from §2.3 Recursion.
/****************************************************************************** * Compilation: javac Permutations.java * Execution: java Permutations n * * Enumerates all permutations on n elements. * Two different approaches are included. * * % java Permutations 3 * abc * acb * bac * bca * cab * cba * ******************************************************************************/ public class Permutations { // print n! permutation of the characters of the string s (in order) public static void perm1(String s) { perm1("", s); } private static void perm1(String prefix, String s) { int n = s.length(); if (n == 0) StdOut.println(prefix); else { for (int i = 0; i < n; i++) perm1(prefix + s.charAt(i), s.substring(0, i) + s.substring(i+1, n)); } } // print n! permutation of the elements of array a (not in order) public static void perm2(String s) { int n = s.length(); char[] a = new char[n]; for (int i = 0; i < n; i++) a[i] = s.charAt(i); perm2(a, n); } private static void perm2(char[] a, int n) { if (n == 1) { StdOut.println(new String(a)); return; } for (int i = 0; i < n; i++) { swap(a, i, n-1); perm2(a, n-1); swap(a, i, n-1); } } // swap the characters at indices i and j private static void swap(char[] a, int i, int j) { char c = a[i]; a[i] = a[j]; a[j] = c; } public static void main(String[] args) { int n = Integer.parseInt(args[0]); String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; String elements = alphabet.substring(0, n); perm1(elements); StdOut.println(); perm2(elements); } }